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Consider the additive group of integer sequences $\mathbb{Z}^{\mathbb{N}}$. Why does every epimorphism of groups $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ split? $(\star)$

Actually this claim is equivalent to the Whitehead problem for countable abelian groups:

"$\Rightarrow$": Recall Specker's result which states that $\mathbb{Z}^{(\mathbb{N})} \to \hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$, $e_i \mapsto \mathrm{pr}_i$ is an isomorphism, i.e. $\mathbb{Z}^{\oplus \mathbb{N}}$ is reflexive. Now assume that $A$ is countable and $\mathrm{Ext}^1(A,\mathbb{Z})=0$. Choose a presentation $0 \to P \to Q \to A \to 0$ with free abelian groups $P,Q$, w.l.o.g. of rank $\aleph_0$. By assumption $Q^* \to P^*$ is an epi, hence splits. Since $P,Q$ are reflexive, then also $P \to Q$ splits, and $A$ is free.

"$\Leftarrow$": If $f : \mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ is an epimorphism, then $f^* : \mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\oplus \mathbb{N}}$ is a monomorphism, the cokernel $A$ of $f^*$ is countable and satisfies $\mathrm{Ext}^1(A,\mathbb{Z})=0$, since $f^{**} \cong f$ is epi. Hence $A$ is free, which implies that $f^*$ splits and therefore also $f^{**} \cong f$ splits. $~\square$

The countable Whitehead problem was proved by Stein in 1950. He used injective resolutions, i.e. $\mathrm{Ext}^1(A,\mathbb{Z}) = \mathrm{coker}(\hom(A,\mathbb{Q}) \to \hom(A,\mathbb{Q}/\mathbb{Z}))$. In particular, $(\star)$ is true. On the other hand, the equivalence above suggests an alternative proof of the countable Whitehead problem. Therefore my question is: Is there a direct proof for $(\star)$?

By Specker's result an endomorphism of $\mathbb{Z}^{\mathbb{N}}$ corresponds to an endomorphism of $\mathbb{Z}^{\oplus \mathbb{N}}$, and therefore to a column-finite matrix. But I don't know how to characterize surjectivity. And this is where I get stuck.

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Do you happen to know it's true? –  Fernando Muro Aug 15 '13 at 9:50
    
@Fernando: The Whitehead problem was proved for countable groups. Martin is merely trying an equivalent approach in proving it. –  Asaf Karagila Aug 15 '13 at 9:56
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Martin, why is it so? –  Fernando Muro Aug 15 '13 at 10:24
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@MartinBrandenburg - apologies - I misread $\mathbb{N}$ for $N$! –  HJRW Aug 15 '13 at 15:01
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In view of @HJRW comment I cannot resist but mentioning that this is one more reason not to rely that much on formulas and formatting details but to describe things verbally if possible. Not sure if this was the case here, but on a mobile device I use that renders MathJax well in general mathbb does not display as such but in a 'normal' font. –  quid Aug 15 '13 at 15:31
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